Understanding Capacitors in Series

When capacitors are connected in series, the overall effective capacitance is less than the capacitance of any individual capacitor. This is because the charge distribution effectively spreads across multiple plates.

Formula for Series Capacitance

The formula used to calculate the total effective capacitance ($C_{total}$) for capacitors connected in series is the reciprocal of the sum of the reciprocals of the individual capacitances ($C_1, C_2, ...$). For two capacitors ($C_1$ and $C_2$) in series, the formula is:

$$ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} $$

Alternatively, this can be rearranged to:

$$ C_{total} = \frac{C_1 \times C_2}{C_1 + C_2} $$

Calculating Total Effective Capacitance

In this problem, we have two capacitors, each with a capacitance of 0.25 μF, connected in series. Let:

• $C_1 = 0.25 \, \mu\text{F}$
• $C_2 = 0.25 \, \mu\text{F}$

Using the formula for two capacitors in series:

$$ C_{total} = \frac{0.25 \, \mu\text{F} \times 0.25 \, \mu\text{F}}{0.25 \, \mu\text{F} + 0.25 \, \mu\text{F}} $$

First, calculate the product of the capacitances in the numerator:

$$ 0.25 \times 0.25 = 0.0625 $$

Next, calculate the sum of the capacitances in the denominator:

$$ 0.25 + 0.25 = 0.50 $$

Now, divide the numerator by the denominator:

$$ C_{total} = \frac{0.0625 \, (\mu\text{F})^2}{0.50 \, \mu\text{F}} $$

$$ C_{total} = 0.125 \, \mu\text{F} $$

Comparing with Options

The calculated total effective capacitance is 0.125 μF. Let's compare this with the given options:

The calculated value matches Option 4.